Problem: Suppose we have a vector-valued function $g(t)$ and a scalar function $f(x, y, z)$. Let $h(t) = f(g(t))$. We know: $\begin{aligned} &g(-2) = (1, 2, 0) \\ \\ &g'(-2) = (0, 4, 3) \\ \\ &\nabla f(1, 2, 0) = (-10, 2, -3) \end{aligned}$ Evaluate $\dfrac{d h}{d t}$ at $t = -2$. $h'(-2)=$
Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(-2) = \nabla f(g(-2)) \cdot g'(-2)$. We know the following. $\begin{aligned} &g(-2) = (1, 2, 0) \\ \\ &g'(-2) = (0, 4, 3) \\ \\ &\nabla f(1, 2, 0) = (-10, 2, -3) \end{aligned}$ Substituting: $h'(-2) = (-10, 2, -3) \cdot (0, 4, 3) = -1$ Answer Therefore, $h'(-2) = -1$.